Question

B1, tính:
a, (x-4).(x+4) 
b, (x-5).(x+5)
B2, viết các đa thức sau dưới dạng bình phương của 1 tổng:
a, x^2 - 2x + 1
b, x^2 + 2x + 1
c, x^2 - 6x + 9
  
MÌNH CẦN GIẢI CHI TIẾT NHÉ< THKS >

Answer 1

B1: 

a) \(\left(x-4\right)\left(x+4\right)=x^2-16\)

b) \(\left(x-5\right)\left(x+5\right)=x^2-25\)

B2:

a) \(x^2-2x+1=\left(x-1\right)^2\)

b) \(x^2+2x+1=\left(x+1\right)^2\)

c) \(x^2-6x+9=\left(x-3\right)^2\)

Answer 2

Bài 1 :

a) \(\left(x-4\right)\left(x+4\right)=x^2-4x+4-16=x^2-16\)

b) \(\left(x-5\right)\left(x+5\right)=x^2-5x+5x-25=x^2-25\)

Bài 2 :

a) \(x^2+2x+1=x^2-x-x+1\)

\(=x.\left(x-1\right)-\left(x+1\right)=\left(x-1\right)^2\)

b) \(x^2+2x+1=x^2+x+x+1\)

\(=x\left(x+1\right)+\left(x+1\right)=\left(x+1\right)^2\)

c) \(x^2-6x+9=x^2-3x-3x+9\)

\(=x.\left(x-3\right)-3.\left(x-3\right)=\left(x-3\right)^2\)

Answer 3

Bài 1 : 

\(\left(x-4\right)\left(x+4\right)=x^2+4x-4x-16=x^2-16\)

\(\left(x-5\right)\left(x+5\right)=x^2+5x-5x-25=x^2-25\)

Bài 2 : 

\(x^2-2x+1=\left(x-1\right)^2\)

\(x^2+2x+1=\left(x+1\right)^2\)

\(x^2-6x+9=\left(x-3\right)^2\)

Answer 4

               Trả lời

Bài 1 : 

\(\left(x-4\right)\left(x+4\right)=x^2+4x-4x-16=x^2-16\)

\(\left(x-5\right)\left(x+5\right)=x^2+5x-5x-25=x^2-25\)

Bài 2 : 

\(x^2-2x+1=\left(x-1\right)^2\)

\(x^2+2x+1=\left(x+1\right)^2\)

\(x^2-6x+9=\left(x-3\right)^2\)

Answer 5

B1 

a, 

\(=x^2-4^2\)  

\(=x^2-16\) 

b, 

\(=x^2-5^2\)   

\(=x^2-25\)                          

B2 : 

a, 

\(=x^2-2\cdot x\cdot1+1^2\)  

\(=\left(x-1\right)^2\)  

b, 

\(=x^2+2\cdot x\cdot1+1^2\)     

\(=\left(x+1\right)^2\)    

c, 

\(=x^2-2\cdot x\cdot3+3^2\)    

\(=\left(x-3\right)^2\)